Given a binary tree, return the preorder traversal of its nodes' values.
GeeksforGeeks
ProgramCreek
YouTube
Input: [1, null, 2, 3]
Output: [1, 2, 3]
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// Definition for a binary tree node.
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) { this.val = val; }
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}/**
* Time complexity :
* Space complexity :
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> nodes = new ArrayList();
helper(root, nodes);
return nodes;
}
private void helper(TreeNode node, List<Integer> nodes) {
if(node == null) return;
nodes.add(node.val);
helper(node.left, nodes);
helper(node.right, nodes);
}
}/**
* Time complexity :
* Space complexity :
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList();
LinkedList<TreeNode> stack = new LinkedList();
stack.add(root);
while(!stack.isEmpty()) {
TreeNode tmp = stack.pollLast();
result.add(tmp.val);
if(tmp.right != null) {
stack.add(tmp.right);
}
if(tmp.left != null) {
stack.add(tmp.left);
}
}
return result;
}
}